Given two words (start and end), and a dictionary, find the length of shortest transformation sequence from start to end, such that:
- Only one letter can be changed at a time
- Each intermediate word must exist in the dictionary
For example,
Given:
start = "hit"
end = "cog"
dict = ["hot","dot","dog","lot","log"]
As one shortest transformation is "hit" -> "hot" -> "dot" -> "dog" -> "cog",
return its length 5.
Note:
- Return 0 if there is no such transformation sequence.
- All words have the same length.
- All words contain only lowercase alphabetic characters.
class Solution {
public:
int ladderLength(std::string start, std::string end, std::unordered_set<std::string> &dict) {
std::unordered_map<std::string, int> dis;
std::queue<std::string> q;
dis[start] = 1;
q.push(start);
while (!q.empty()) {
std::string word = q.front(); q.pop();
if (word == end) break;
for (int i = 0; i < word.size(); i++) {
for (int j = 0; j < 26; j++) {
std::string newWord = word;
newWord[i] = 'a' + j;
if (dict.count(newWord) > 0 && dis.count(newWord) == 0) {
dis[newWord] = dis[word] + 1;
q.push(newWord);
}
}
}
}
if (dis.count(end) == 0) return 0;
return dis[end];
}
};
本文章为转载内容，我们尊重原作者对文章享有的著作权。如有内容错误或侵权问题，欢迎原作者联系我们进行内容更正或删除文章。
